$\frac{xyz}w+\frac{yzw}x+\frac{zwx}y+\frac{wxy}z$$\geq^4\sqrt{\frac{x^3y^3z^3w^3}{xyzw}}$$=x^2y^2z^2w^2$ Therefore $x^2y^2z^2w^2\leq4$, which means that $|xyzw|\leq2$. Since $x$, $y$, $z$, $w$ are all nonzero, we would no that either the absolute value of all the four numbers are 1, or there is a number which has an absolute value of 2. In the first occasion, There can be four 1, two 1 and two -1 or four -1. As for the latter, It's obvious that it wouldn't work. Which means that the solution to the problem is (x, y, z, w) = (1, 1, 1, 1), (1, 1, -1, -1), (1, -1, 1, -1), (-1, 1, 1, -1), (1, -1, -1, 1), (-1, 1, -1, 1), (-1, -1, 1, 1), (-1, -1, -1, -1).